H
ello,
Please reply to the list. You will have more chance to get a reply.
As I am not a Windows user, I can't help you more. But this problem has
been reported in this thread, with a reply by Paul Murrell, one of the
author of the package you are trying to use.
Hi,
There are many occurrences of the CIK number in the page source. This pulls
out the first node containing it:
node - getNodeSet(doc[[1]], //link[@rel='alternate'] )
From there you can extract the number. Here's one way to do it.
strsplit(strsplit(unlist(node)[[5]], CIK=)[[1]][2],
Hi,
Is the lda function (R MASS package) “Proportion of trace” is similar to
“proportion of variance explained”in the case of PCA?
How can I store the LD1 and LD2 in two separate variables?
Thanks
--
View this message in context:
Hallo.
I'm running step() on lm() in R version 2.15.1 within RStudio on my
xUbuntu Linux computer with 4 core processor.
Linux top command states that rsession (R) is using from 100 % to
about 330 % of the processor, which (I think) means that R is using
more than one core. Is it possible? I
I am have a procedure which generates multidimensional arrays.
To compute them is expensive so I want to store them in order to be
able to analyse them later.
I am using at the moment
problem is that the array is always assigned to a variable ma (the
computation is in a loop).
Than I generate a
On Tue, 13 Aug 2013, Cade, Brian wrote:
Lauria: For historical reasons the logistic regression (binomial with
logit link) model portion of a zero-inflated count model is usually
structured to predict the probability of the 0 counts rather than the
nonzero (=1) counts so the coefficients will
Thank you for your answer Ben B., it is helpful.
The post on Fixed vs Random effects is particularly interesting.
I had in mind to create a fixed interaction, as you propose (time*A). I
actually wanted to compare it with a random interaction, so to decide on which
model to go with based on
On Tue, 13 Aug 2013, Walter Anderson wrote:
I have a set of survey data where I have answers to identify preference of
three categories using three questions
1) a or b?
2) b or c?
3) a or c?
and want to obtain weights for each of the preferences
something like X(a) + Y(b) + Z(c) = 100%
Hi
is there a way to generate a pdf, e.g. pdf(test.pdf), even if the file
test.pdf is open? (e.g. with acrobat)
i.e. I would have to close test.pdf in the viewer without any user
interaction, than override the file and maybe open it again.
thx
Christof
Ambiguity is indeed detected by R and the user is informed on it. But in the
case of Hadley's example, I still believe, that the specific multiple
inheritance structure creates this behavior. If you call:
showMethods(f)
Function: f (package .GlobalEnv)
x=A, y=A
x=AB, y=AB
(inherited from:
You can use saveRDS() and readRDS() instead of save/load. Unlike load
readRDS just returns the value, allowing you to assign it to whatever name
you want.
Best,
Ista
On Wed, Aug 14, 2013 at 5:52 AM, Witold E Wolski wewol...@gmail.com wrote:
I am have a procedure which generates
Based on the plot of Schoenfeld residuals and Terry's explanation is it safe
to say that proportional hazards assumption holds despite the significant
global p-values?
No. I don't want to put words in Terry's mouth, but he seems to be saying that
proportional hazards does NOT hold but it may
Dear Robert,
(1|A/B) is shorthand for (1|A) + (1|A:B)
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43
On 14/08/2013 11:23, Christof Kluß wrote:
Hi
is there a way to generate a pdf, e.g. pdf(test.pdf), even if the file
test.pdf is open? (e.g. with acrobat)
i.e. I would have to close test.pdf in the viewer without any user
interaction, than override the file and maybe open it again.
thx
I have a string that contains something like:
...verified email at neu.edubrCited by 99853br/td/tr/table/
div/div/div/divdiv...
and I'd like to extract the number next to the text Cited by - so
it will be whatever numbers are beside Cited by until a non-numeric
character is reached.
Hi,
Try:
a1- ...verified email at neu.edubrCited by
99853br/td/tr/table/div/div/div/divdiv...
gsub(.*Cited by\\s+([0-9]+).*,\\1,a1)
#[1] 99853
A.K.
- Original Message -
From: Thomas thomas.ches...@nottingham.ac.uk
To: r-help@r-project.org
Cc:
Sent: Wednesday, August 14, 2013 8:14 AM
Thomas,
se ?sub and regular expression. That should make it. Further, see the package
gsubfn
Best
Simon
On Aug 14, 2013, at 2:14 PM, Thomas thomas.ches...@nottingham.ac.uk wrote:
I have a string that contains something like:
...verified email at neu.edubrCited by
Thanks, I will look into ways to tell ess/emacs to use such options. I
am, however, quite sure that I have never answered yes to the question
when quitting R.
Cheers
Jannis
On 13.08.2013 20:21, MacQueen, Don wrote:
R --no-save
__
Hello all,
I've tried to solve this for weeks and posted to other forums with
little success- I'd appreciate any help from anyone.
I have survey data grouped by facility and area (area is a collection of
facilities). Questions are q1-q10.
For each facility, I need to subset each item into
Thank you very much Kevin Wright! you managed to understand my somewhat cryptic
question.
Now I have a perfect graph. I changed the my.theme to:
my.theme - list(
box.umbrella = list(col = black),
box.rectangle = list(col=black),
box.dot = list(col = black, pch = 3, cex=2),
plot.symbol
Hi A.K,
Thanks for your great help.
I'm now running your first suggestion on a 600.000 row sample after
verifying it works on a smaller sample.
It's now been running for 40 minutes.
Which method do you think will be faster?
Regards Derk
--
View this message in context:
Hi,
I'm attempting to make a bar plot for some genomics data that includes
a separate bar for each sample taken. I am having trouble applying these sample
labels to the individual bars. It seems that the barplot() function will only
take a numeric matrix, and therefore cannot have any
Hi,
Your question doesn't quite make sense to me, and since you didn't
provide a reproducible example it's impossible to really know what
you're doing.
Have you read ?barplot and tried the examples? There's a names.arg
argument that takes a character vector that is used for labels.
Otherwise the
Hi Again,
I still cannot get the PostscriptTrace to work. The error message remains
the same.
I have run the following with no problems
Sys.setenv(R_GSCMD=c:/Program Files(x86)/gs/gs9.07/bin/gswin32)
but the call to Postscript Trace still fails
PostScriptTrace(fish.ps)
Error in
Well, I have made some tests with the 'no save' option. This only seems
to control the saving of .RData files (at least none appeared in the
working directory in my tests). A file called .RHistory is still
created. I can now put some code to delete this file in .Last function
but somehow I
Hi,
Try:
set.seed(42)
dat1- as.data.frame(matrix(sample(LETTERS,2*1e6,replace=TRUE),ncol=2))
dat2- dat1[1:1e5,]
dat3- dat1
library(data.table)
dt1- data.table(dat1)
system.time(dat1$sat- 1*(dat1[,1]==dat1[,2]))
# user system elapsed
# 0.148 0.004 0.152
library(car)
system.time({dat3$sat-
OK, this seems to be only a problem when I use emacs/ess. I will try to
find a solution to this but this does not seem to be related to emacs.
In case anyone of you has an Idea: I use:
(setq inferior-R-args --no-save --no-restore --silent)
to start R, but still a .RHistory file is saved.
On 14/08/2013 14:52, Andrew Halford wrote:
Hi Again,
I still cannot get the PostscriptTrace to work. The error message remains
the same.
I have run the following with no problems
Sys.setenv(R_GSCMD=c:/Program Files(x86)/gs/gs9.07/bin/gswin32)
but the call to Postscript Trace still fails
Dear R forum,
I have a function which generates say two outputs, say output_1 and output_2.
Output_1 is a single row output whereas Output_2 is a dataframe having multiple
records. Is it possible to use two return statements in function. Output_2 uses
some records from output_1, hence I need
Hi,
The usual thing to do is return a list containing output_1 and output_2.
Sarah
On Wed, Aug 14, 2013 at 10:23 AM, Katherine Gobin
katherine_go...@yahoo.com wrote:
Dear R forum,
I have a function which generates say two outputs, say output_1 and output_2.
Output_1 is a single row output
Dear Katherine,
Combine both outputs in a list and return that.
return(list(first = output.1, second = output.2))
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality
Hi,
I tried the second method on a bigger dataset. This is what I get,
indx-rep(1:nrow(dat1),6e4)
dat2- dat1[indx,]
system.time({
vec1- paste(dat2[,1],dat2[,2],colnames(dat2)[2],sep=.)
res2-reshape(dat2,idvar=newCol,varying=list(2:26),direction=long)
res3-res2[order(res2[,4]),]
res4-
I am learning analysis of longitudinal data from the book
Longitudinal data Data Analysis for the Behavioral Science using R by Jeffrey
Long.
On page 123,
myX - scale_x_continuous(breaks = 5:8, name = Grade)
myY - scale_y_continuous(name = Reading Score )
g1 - ggplot(data=MPLS.LS,
I'm not trying to repost or spam everyone- I submitted this once before
I subscribed so I just wanted to resend in case it didn't get
disseminated.
Hello all,
I've tried to solve this for weeks and posted to other forums with
little success- I'd appreciate any help from anyone.
I
I'm sure there are better, more elegant ways avoiding the nested loop I'm
suggesting - but if it was my problem, here is what I would do (assuming that
my understanding of your question is correct):
### separate function for 'doing something' with the data subset
do.something - function( qA, qB
Carlos,
There are likely several problems with your likelihood. You should check it
carefully first before you do any optimization. It seems to me that you have
box constraints on the parameters. They way you are enforcing them is not
correct. I would prefer to use an optimization algorithm
Z is correct, of course. I was just being a little too simplistic in my
explanation trying to emphasize the reversal of signs of the coefficients
in the logistic regression part of the zero-inflated model.
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre
A look at ?history shows an environment variable that might help you
restrict it to just one central .Rhistory file.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 8/14/13 7:15 AM, Jannis bt_jan...@yahoo.de wrote:
OK,
In my opinion the reason for the behavior lies in the specific multiple
inheritance structure between AB, B and A.
So what if we don't make such a weird inheritance structure, and
instead have A and B inherit from a common parent:
setClass(A, contains = list)
setClass(B, contains = list)
From: Kevin E. Thorpe [mailto:kevin.tho...@utoronto.ca]
Sent: Tuesday, August 13, 2013 2:25 PM
Subject: Re: [R] Memory limit on Linux?
It appears that at the shell level, the differences are not to blame.
It has been a long time, but years ago in HP-UX, we needed to change an
actual
If your efforts on those unspecified other forums were like this one, perhaps
repeating what doesn't work is your problem. You don't appear to have read and
understood the footer of any email on this list. For one thing, you need to
post in plain text (to avoid the corruption that HTML email
From: Jack Challen [mailto:jack.chal...@ocsl.co.uk]
Sent: Wednesday, August 14, 2013 10:45 AM
Subject: RE: Memory limit on Linux?
(I'm replying from a horrific WebMail UI. I've attempted to maintain
what I think is sensible quoting. Hopefully it reads ok).
[snip]
If all users are able to
Hello again,
I need to calculate the week number of the corresponding month given a date.
Is there any function available with R to calculate that?
Thanks and regards,
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Because the signature is always (A,A) or (B,B). Then, as in AB we have A and B
and no relationship between A and B, R chooses the method lexicographically.
The result is as expected: f for A is chosen.
If you would do something like:
setClass(A, contains = list)
setClass(B, contains = list)
mod Jeff Newmiller's comments...
1. Have you readAn Introduction to R? (or other basic tutorial --
there are many on the web). If no, stop posting and do so. This will
help you to understand R's basic data manipulation capabilities and
structures (list, apply type functions,...).
2. mod 1),
What about
?lubridate
particulatrly week()?
On Wednesday 14 August 2013 22:00:42 Christofer Bogaso wrote:
Hello again,
I need to calculate the week number of the corresponding month given a date.
Is there any function available with R to calculate that?
Thanks and regards,
On Wed, Aug 14, 2013 at 11:36 AM, Simon Zehnder szehn...@uni-bonn.de wrote:
Because the signature is always (A,A) or (B,B). Then, as in AB we have A and
B and no relationship between A and B, R chooses the method
lexicographically. The result is as expected: f for A is chosen.
It's not as
Hi Rainer,
Thanks for your pointer. However I was not looking the week number for the
underlying year. Rather, week number for the underlying month.
For example, if the date is 8/14/2013 then I should get the week number as
'3'. Because it is 3rd week of August month.
Any help on that?
Thanks
Is there an easy way to convert character strings with comma-separated
numbers and ranges to a numeric vector?
x- 2,5-7,10,12-15
[1] 2 5 6 7 10 12 13 14 15
Thanks,
Chris
--
Chris Stubben
Los Alamos National Lab
Bioscience Division
MS M888
Los Alamos, NM 87545
Hi Zehnder,
You're right that the fact that B already inherits from A is probably
part of the story but it's not all the story:
setClass(A, NULL)
setClass(B, A)
setMethod(show, A, function(object) cat(A object\n))
setMethod(show, B, function(object) cat(B object\n))
setClass(C, B)
Hi,
This seemed to be faster than the other two methods:
vec1- as.character(rep(dat1[,1],each=(ncol(dat1)-1)))
vec2- as.character(unlist(t(dat1[,-1])))
vec3- rep(rep(c(TRUE,FALSE),c(1,(ncol(dat1)-2))),nrow(dat1))
dat2-data.frame(DSYSRTKY=vec1,CODE=vec2,PRIMAIRY=vec3,stringsAsFactors=FALSE)
dat3-
On Aug 14, 2013, at 12:41 PM, Chris Stubben stub...@lanl.gov wrote:
Is there an easy way to convert character strings with comma-separated
numbers and ranges to a numeric vector?
x- 2,5-7,10,12-15
[1] 2 5 6 7 10 12 13 14 15
Thanks,
Chris
There is a general admonishment to
Hello,
I am trying to make a slight modification to the way gam (mgcv) works.
I want to modify the G$X matrix, which is the design matrix, to
accommodate an estimator that I am trying to program. I am working with
panel data, and I want to take all continuous variables, including basis
Dear All,
I plan to switch to linux and run R, latex and CUDA on it.
For this reason, which is the recommended version of linux (stable,
efficient, compatible etc.)?
It will be installed on a desktop, with intel i5, nvidia gpu, 16 gb ram.
Thank for your attention and hopefully not off topic.
Howdy,
On Wed, Aug 14, 2013 at 9:40 AM, Bert Gunter gunter.ber...@gene.com wrote:
mod Jeff Newmiller's comments...
1. Have you readAn Introduction to R? (or other basic tutorial --
there are many on the web). If no, stop posting and do so. This will
help you to understand R's basic data
The one you can have some assistance from friends an colleagues around.
Ditoni + tastierina =erroru di battitura
Jie jimmycl...@gmail.com ha scritto:
Dear All,
I plan to switch to linux and run R, latex and CUDA on it.
For this reason, which is the recommended version of linux (stable,
And it doesn't even select the first method lexicographically in the
ordering (whatever that means):
setClass(B, NULL)
setClass(A, NULL)
setMethod(show, B, function(object) cat(B object\n))
setMethod(show, A, function(object) cat(A object\n))
setClass(AB, contains=c(B, A))
ab -
Hi,
You could try:
dat1- read.table(text=
X1,X2,X3
age,race,stat
12,2,35
17,6,55
,sep=,,header=TRUE,stringsAsFactors=FALSE)
colnames(dat1)- dat1[1,]
dat1- dat1[-1,]
dat1[]-lapply(dat1,as.numeric)
row.names(dat1)- 1:nrow(dat1)
dat1
# age race stat
#1 12 2 35
#2 17 6 55
A.K.
Dear all,
I need to insert an small icon in each bar of a barplot, in a specific
location (depending on bar's value).
For example: the first bar has X value of 5. I need to insert an icon at X
value 3. The second bar has X value of 8. I need to insert an icon at value
7.
I have both vectors with
Hello,
I'm using igraph for some simple network analysis of a directed graph.
When I calculate the page-rank using page.rank(g), the returned vector
contains negative values. The vector sums to 1, which tells me that the
algorithm working fine.
How do I interpret the negative values and is
(This is a repost from a little while ago. I assume my mail got silently
bounced because I used some rather strange email routing. If it did get
through, and I simply haven't seen it or a response, then please accept my
apologies)
Hi,
I'm new to R, and new to statistics. I'm *trying* to learn
Hi Arun,
The second method is indeed working much faster. It worked fast for my
600.000 row record.
Still I have 2 bigger files where processing becomes an issue even though I
have lots of memory (32 gig) for the second statement:
res2-reshape(dat2,idvar=newCol,varying=list(2:26),direction=long)
Dear all,
I am exploring ways to perform multiplication of a 9 x 4 matrix with
it's transpose.
As expected even a 4 x 100 %*% 100x4 didn't work on my desktop...
giving the error Error: cannot allocate vector of length 16
However I am trying to run this on one node (64GB
Dear Brian and Achim,
Many thanks for your reply and help it is very much appreciated!
All the best,
Valentina
Dr. Valentina Lauria
Postdoctoral researcher
Room 118, Martin Ryan Institute
Department of Earth and Ocean Sciences
National University of Ireland, Galway
Ireland
Do your matrices have any special properties we should know about? For example,
are they sparse, symmetric, diagonal, etc?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Praveen Surendran
Sent: Wednesday, August 14, 2013 11:41 AM
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of arun
Sent: Wednesday, August 14, 2013 11:56 AM
To: farnoosh sheikhi
Cc: R help
Subject: Re: [R] changing colnames
Hi,
You could try:
dat1- read.table(text=
X1,X2,X3
On Aug 14, 2013, at 11:08 AM, Andrew Crane-Droesch wrote:
Hello,
I am trying to make a slight modification to the way gam (mgcv) works.
I want to modify the G$X matrix, which is the design matrix, to
accommodate an estimator that I am trying to program. I am working with
panel data,
On Wed, 14 Aug 2013, Cade, Brian wrote:
Z is correct, of course. I was just being a little too simplistic in my
explanation trying to emphasize the reversal of signs of the
coefficients in the logistic regression part of the zero-inflated model.
When users ask me what the binary part of the
Thanks! I hadn't touched environment variables before, not knowing what
they were. The G$X matrix indeed seems to get squared, giving
appropriately nonsensical results. I can now go and code a (hopefully)
sensible change to it.
On 08/14/2013 10:32 PM, David Winsemius wrote:
Well, one might start by noting that it's usually unnecessary and
unwise to multiply a matrix by its transpose.
Matrix decompositions, algebraic identities, and/or iterative
procedures usually do calculations involving t(x)%*%x in better ways.
'Course without knowledge of your problem, maybe I'm
On Aug 14, 2013, at 12:46 PM, Andrew Crane-Droesch wrote:
Thanks! I hadn't touched environment variables before, not knowing what they
were. The G$X matrix indeed seems to get squared, giving appropriately
nonsensical results. I can now go and code a (hopefully) sensible change to
it.
Hello,
I have some stock pricing data for one minute intervals.
The delivery format is a bit odd. The date column is easily parsed and used as
an index for an its object. However, the time column is just an integer
(1:1807)
I just need to extract the *last* entry for each day. Don't
On Wed, 14-Aug-2013 at 04:01PM +0200, Jannis wrote:
| Well, I have made some tests with the 'no save' option. This only
| seems to control the saving of .RData files (at least none appeared
| in the working directory in my tests). A file called .RHistory is
| still created. I can now put some
Hi,
Try:
dat1- read.table(text=
Date Time O H L C U D
06/01/2010 1358 136.40 136.40 136.35 136.35 2 12
06/01/2010 1359 136.40 136.50 136.35 136.50 9 6
06/01/2010 1400 136.45 136.55 136.35 136.40 8 7
06/01/2010 1700 136.55 136.55 136.55 136.55 1 0
06/02/2010
Hi,
this is not a problem of R it is a problem of the pdf viewer.
Solution: just use an alternative pdf viewer like gsview or (even
simpler) SumatraPDF.
Hope it helps
Thomas P.
__
R-help@r-project.org mailing list
That works beautifully.
Never used the unlist or with commands before. More to learn there.
Thanks!
--
Noah Silverman, M.S., C.Phil
UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095
On Aug 14, 2013, at 1:08 PM, arun smartpink...@yahoo.com wrote:
Hi,
Try:
A somewhat faster version (for datasets with lots of dates, assuming it is
sorted by date and time) is
isLastInRun - function(x) c(x[-1] != x[-length(x)], TRUE)
f3 - function(dataFrame) {
dataFrame[ isLastInRun(dataFrame$Date), ]
}
where your two suggestions, as functions, are
f1 -
I'm not sure I follow you exactly so let's start with some data
and one graph and move on from there:
First the data (I'm assuming you don't have A to A so you really
want 3 lines on a graph)?
set.seed(42)
pairs - structure(list(From = structure(c(1L, 1L, 1L, 2L, 2L,
2L, 3L,
3L, 3L, 4L, 4L,
While we're playing code golf, likely faster still could be to use
data.table. Assume your data is in a data.frame named x:
R library(data.table)
R x - data.table(x, key=c('Date', 'Time'))
R ans - x[, .SD[.N], by='Date']
-steve
On Wed, Aug 14, 2013 at 2:01 PM, William Dunlap wdun...@tibco.com
Or with plyr:
R library(plyr)
R ans - ddply(x, .(Date), function(df) df[which.max(df$Time),])
-steve
On Wed, Aug 14, 2013 at 2:18 PM, Steve Lianoglou
lianoglou.st...@gene.com wrote:
While we're playing code golf, likely faster still could be to use
data.table. Assume your data is in a
On Aug 14, 2013, at 8:16 AM, MacQueen, Don wrote:
A look at ?history shows an environment variable that might help you
restrict it to just one central .Rhistory file.
I'm guessing this refers to the fourth paragraph and it appears that
suppressing any history saving may be possible as well.
Is there a more concise way to write the following code?
library(gdata)
mydataOUTPUTrtfA - read.csv(mergedStatstA.csv)
save(mydataOUTPUTrtfA, file=mydataOUTPUTrtfA.RData)
mydataOUTPUTrtfA - rename.vars(mydataOUTPUTrtfA, from=X,
to=Statistics.Calculated, info=FALSE)
mydataOUTPUTrtfB -
Or how about rle/cumsum, as per the appended?
-- Mike
myData - read.table(junk.dat, header=TRUE, stringsAsFactors=FALSE)
myData
Date Time O H L C U D
1 06/01/2010 1358 136.40 136.40 136.35 136.35 2 12
2 06/01/2010 1359 136.40 136.50 136.35 136.50 9 6
3 06/01/2010 1400 136.45 136.55 136.35
On Aug 14, 2013, at 2:18 PM, Steve Lianoglou wrote:
While we're playing code golf, likely faster still could be to use
data.table. Assume your data is in a data.frame named x:
R library(data.table)
R x - data.table(x, key=c('Date', 'Time'))
R ans - x[, .SD[.N], by='Date']
I though
Hi,
May be this helps:
library(gsubfn)
as.numeric(strsplit(gsub([c() ],,gsubfn(([0-9]+)-([0-9]+),
~as.numeric(seq(x,y)),x)),,)[[1]])
#[1] 2 5 6 7 10 12 13 14 15
A.K.
- Original Message -
From: Chris Stubben stub...@lanl.gov
To: r-help@r-project.org
Cc:
Sent: Wednesday, August 14,
In the event that these are moderately sparse matrices, you could try Matrix or
SparseM.
Roger Koenker
rkoen...@illinois.edu
On Aug 14, 2013, at 10:40 AM, Praveen Surendran wrote:
Dear all,
I am exploring ways to perform multiplication of a 9 x 4 matrix with
it's transpose.
Neither gsubfn nor eval(parse)) is required, of course:
x- 2,5-7,10,12-15
z -strsplit(scan(text=x,sep=,,wh=a),split=-) ## use scan to vectorize
the string
l - lapply(z,as.numeric)
unlist(lapply(l,function(x){
last - x[length(x)]## alternately could use if() on the length of x
On 08/14/2013 10:40 PM, Alexander Gotowski wrote:
Hi,
I'm attempting to make a bar plot for some genomics data that includes
a separate bar for each sample taken. I am having trouble applying these sample
labels to the individual bars. It seems that the barplot() function will only
I would use source
x- 2,5-7,10,12-15
source(textConnection(paste(c(, gsub(\\-, :, x), $value
[1] 2 5 6 7 10 12 13 14 15
On Wed, Aug 14, 2013 at 8:10 PM, Bert Gunter gunter.ber...@gene.com wrote:
Neither gsubfn nor eval(parse)) is required, of course:
x- 2,5-7,10,12-15
z
On 08/14/2013 03:43 PM, bcrombie wrote:
Is there a more concise way to write the following code?
library(gdata)
mydataOUTPUTrtfA - read.csv(mergedStatstA.csv)
save(mydataOUTPUTrtfA, file=mydataOUTPUTrtfA.RData)
mydataOUTPUTrtfA - rename.vars(mydataOUTPUTrtfA, from=X,
to=Statistics.Calculated,
On 08/15/2013 01:16 AM, Igor Ribeiro wrote:
Dear all,
I need to insert an small icon in each bar of a barplot, in a specific
location (depending on bar's value).
For example: the first bar has X value of 5. I need to insert an icon at X
value 3. The second bar has X value of 8. I need to insert
I had earlier came up with a similar kind of function, though didn't posted.
unlist(lapply(strsplit(x,,)[[1]],function(x)
sapply(strsplit(x,-),function(x) {x1-as.numeric(x);if(length(x1)==2)
seq(x1[1],x1[2]) else x1})))
A.K.
- Original Message -
From: Bert Gunter
Better yet!
-- Bert
On Wed, Aug 14, 2013 at 6:03 PM, Richard M. Heiberger r...@temple.edu wrote:
I would use source
x- 2,5-7,10,12-15
source(textConnection(paste(c(, gsub(\\-, :, x), $value
[1] 2 5 6 7 10 12 13 14 15
On Wed, Aug 14, 2013 at 8:10 PM, Bert Gunter
HI,
You could try:
#If all the files are in the working directory:
vec1-list.files() #Created 3 dummy files in my WD
vec1
#[1] mergedStatstA.csv mergedStatstB.csv mergedStatstC.csv
library(gdata)
lapply(seq_along(vec1),function(i)
I would like to be able to use gsub or gsubfn to process a formula and
to translate the variables but to ignore expressions in the formula.
Supposing that the R formula has already been transformed into a
character string and that the transformation is to convert variable
names to upper case
This might be hard.
How to tell f is to be changed while h is NOT ...
Thanks,
Guanrao
http://www.myfav5.com
where fun and easy friend-making happens
From: Frank Harrell f.harr...@vanderbilt.edu
To: RHELP r-h...@stat.math.ethz.ch
Sent: Wednesday, August
I run the examples in delayedAssign:
msg - old
delayedAssign(x, msg)
msg - new!
x
If I run these four commands together, x is new. If I run the first two
commands first and then run the last two commands, x is old.
I just cannot figure out why.
Thanks.
Gang
[[alternative HTML version
Thanks for the code. It was so simple and worked perfectly.
I really appreciate it.
Best,Farnoosh Sheikhi
Cc: R help r-help@r-project.org
Sent: Wednesday, August 14, 2013 11:56 AM
Subject: Re: changing colnames
Hi,
You could try:
dat1- read.table(text=
The manual seems to suggest, with the SIMPLIFY = TRUE default option,
Vectorize would conjure a vector if possible.
Quote:
SIMPLIFY: logical or character string; attempt to reduce the result to
a vector, matrix or higher dimensional array; see the
‘simplify’ argument of
Thanks so much for looking into this for me.
Unfortunately, I get an error when I execute your code. Is there a
library that you loaded that I haven't?
require(scrapeR)
require(XML)
require(RCurl)
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